For the reaction:
2 AgCl(s) + H2(g) (0.4 atm) → 2 Ag(s) + 2 H+(0.1 M) + 2 Cl-(0.2 M)
Calculate the emf of the cell at 25 °C.
Given: ΔG° = -43500 J mol-1. [log 10 = 1, 1 F = 96500 C mol-1]
2 AgCl(s) + H2(g) (0.4 atm) → 2 Ag(s) + 2 H+(0.1 M) + 2 Cl-(0.2 M)
Calculate the emf of the cell at 25 °C.
Given: ΔG° = -43500 J mol-1. [log 10 = 1, 1 F = 96500 C mol-1]
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Solution and Explanation
Answer:
Step 1, find E°cell: ΔG° = -nFE°cell, so E°cell = -ΔG°/(nF) = -(-43500)/(2 × 96500) = 43500/193000 = 0.2254 V (about 0.225 V).
Step 2, Nernst equation. The reaction quotient is Q = ([H+]2 [Cl-]2) / p(H2) = ((0.1)2 × (0.2)2) / 0.4 = (0.01 × 0.04) / 0.4 = 0.0004 / 0.4 = 0.001 = 10-3.
Ecell = E°cell - (0.0591/n) log Q = 0.2254 - (0.0591/2)(log 10-3) = 0.2254 - (0.02955)(-3) = 0.2254 + 0.08865 = 0.314 V.
So the emf of the cell at 25 °C is approximately 0.314 V. The positive emf shows the reaction is spontaneous as written.
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